Serious question, how would using rust avoid this? Rust still has reference types in the background, right? Still has a way to put stuff on the heap too? Those are the only 2 requirements for reusing memory bugs
This is a use-after-free, which should be impossible in safe Rust due to the borrow checker. The only way for this to happen would be incorrect unsafe code (still possible, but dramatically reduced code surface to worry about) or a compiler bug. To allocate heap space in safe Rust, you have to use types provided by the language like Box, Rc, Vec, etc. To free that space (in Rust terminology, dropping it by using drop() or letting it go out of scope) you must be the owner of it and there may be current borrows (i.e. no references may exist). Once the variable is droped, the variable is dead so accessing it is a compiler error, and the compiler/std handles freeing the memory.
There’s some extra semantics to some of that but that’s pretty much it. These kind of memory bugs are basically Rust’s raison d’etre - it’s been carefully designed to make most memory bugs impossible without using unsafe. If you’d like more information I’d be happy to provide!
Thanks for the response. Ive heard of rust’s compiler being very smart and checking a ton of stuff. Its good thing it does, but i feel like there are things that can cause this issues rust cant catch. Cant put my finger on it.
What would rust do if you have a class A create something on the heap, and it passes this variable ( by ref ? ) to class B, which saves the value into a private variable in class B. Class A gets out of scope, and would be cleaned up. What it put on the heap would be cleaned up, but class B still has a reference(?) to the value on the heap, no? How would rust handle such a case?
It’s not like C where you have control over when you can make references to data. The compiler will stop you from making references in the cases where a memory bug would be possible.
You use lifetimes to annotate parameters and return values in order to tell the compiler about how long things must last for your function to be valid. You can link a specific input with the output, or explicitly separate them. If you don’t give lifetimes the language uses some basic rules to do it for you. If it can’t, eg it’s ambiguous, then it’s a compile error and you need to do it manually.
It’s one of the harder concepts of rust to explain succinctly. But imagine you had a function that took strA and strB, used strB to find a subsection of strA, and then return a slice of strA. That slice is tied to strA. You would use 'a annotation for strA and the return value, and 'b for strB.
Rust compiler will detect the lifetime being shorter than expected.
Also, ownership semantics. Think c++ move semantics. Only one person is left with a good value, the previous owners just have garbage data they can’t use anymore. If you created a thing on the heap and then gave it away, you wouldn’t have it anymore to free at the end. If you want to have “multiple owners” then you need ref counting and such, which also stops this problem of premature freeing.
Edit: one more thing: reference rules. You can have many read-only references to a thing, or one mutable reference. Unless you’re doing crazy things, the compiler simply won’t let you have references to a thing, and then via one of those references free that thing, thereby invalidating the other references.
Thats interresting, thanks! Stuff for me to look into!
I also think halfway through the conversation i might have given the impression i was talking about pointers, while it was not my intention to do so. That said, the readonly/mutable reference thing is very interresting!
Ill look into what rust does/has that is like the following psuedocode :
DataBaseUser variable1 = GetDataBaseUser(20);
userService.Users.Add(variable1);
variable1 = null; // or free?
[end of function scope here, reference to heap now in list ]
Rust simply doesn’t allow you to have references to data that goes out of scope (unless previously mentioned hoops are jumped through such as an explicitly declared unsafe block). It’s checked at compile time. You will never be able to compile the program.
Rust isn’t C. Rust isn’t C++. The memory-safe-ness of it is also not magic, it’s a series of checks in the compiler.
That sounds odd. That also means that a mapper, command, service,… can never return a class object or entity. Most of the programming world is based on oop o.O
Keep in mind im not talking about the usage of pointers, but reference typed variables.
The way I understand it, it is a bug in C implementation of free() that causes it to do something weird when you call it twice on the same memory. Maybe In Rust you can never call free twice, so you would never come across this bug. But, also Rust probably doesn’t have the same bug.
My point is it seems it is a bug in the underlying implementation of free(), not to be caught by the compiler, and can’t Rust have such errors no matter its superior design?
The way that rust attempts to prevent this class of error is not by making an implementation of free that is safe to call twice, but by making the compiler refuse to compile programs where free could be called twice on a pointer.
Anyway, use after free doesn’t depend on a double free. It just means that the program frees memory but keeps the pointer (which now points at memory that could contain unrelated data at some future point in time) and if someone trying to exploit the program finds a way to induce the program to read or write to that memory they may be able to access data they are not expected to, or write data to be used by a different part of the program that they shouldn’t be able to
Thanks, I understand the problem with using memory after it’s been freed and possibly access it changed by another part of the process. I guess I was confused by the double free explanation I read, which didn’t really say how it could be exploited, but I think you are right it still needs to be accessed later by the original program, which would not happen in Rust.
Not really, the issue is that C/C++ is not memory safe, i.e. it allows you to access memory that has already been freed. Consider the following C++ code:
That will print 10, but the memory where data was defined has been freed, and is no longer in control of the program. Meaning that if something else allocated that memory they can control what my program does.
Consider that on that example above later in the program we do:
user.access_level = *ptr;
If someone manages to get control of that memory between when we freed it and we used it they can make the access_level of the user be whatever they want.
This is a problem with C/C++ allowing you to access memory that has been freed, which is why C/C++ programmers need to be extra careful.
Serious question, how would using rust avoid this? Rust still has reference types in the background, right? Still has a way to put stuff on the heap too? Those are the only 2 requirements for reusing memory bugs
This is a use-after-free, which should be impossible in safe Rust due to the borrow checker. The only way for this to happen would be incorrect unsafe code (still possible, but dramatically reduced code surface to worry about) or a compiler bug. To allocate heap space in safe Rust, you have to use types provided by the language like
Box,Rc,Vec, etc. To free that space (in Rust terminology, dropping it by usingdrop()or letting it go out of scope) you must be the owner of it and there may be current borrows (i.e. no references may exist). Once the variable isdroped, the variable is dead so accessing it is a compiler error, and the compiler/std handles freeing the memory.There’s some extra semantics to some of that but that’s pretty much it. These kind of memory bugs are basically Rust’s raison d’etre - it’s been carefully designed to make most memory bugs impossible without using
unsafe. If you’d like more information I’d be happy to provide!Thanks for the response. Ive heard of rust’s compiler being very smart and checking a ton of stuff. Its good thing it does, but i feel like there are things that can cause this issues rust cant catch. Cant put my finger on it.
What would rust do if you have a class A create something on the heap, and it passes this variable ( by ref ? ) to class B, which saves the value into a private variable in class B. Class A gets out of scope, and would be cleaned up. What it put on the heap would be cleaned up, but class B still has a reference(?) to the value on the heap, no? How would rust handle such a case?
It’s not like C where you have control over when you can make references to data. The compiler will stop you from making references in the cases where a memory bug would be possible.
You use lifetimes to annotate parameters and return values in order to tell the compiler about how long things must last for your function to be valid. You can link a specific input with the output, or explicitly separate them. If you don’t give lifetimes the language uses some basic rules to do it for you. If it can’t, eg it’s ambiguous, then it’s a compile error and you need to do it manually.
It’s one of the harder concepts of rust to explain succinctly. But imagine you had a function that took strA and strB, used strB to find a subsection of strA, and then return a slice of strA. That slice is tied to strA. You would use
'aannotation for strA and the return value, and'bfor strB.Rust compiler will detect the lifetime being shorter than expected.
Also, ownership semantics. Think c++ move semantics. Only one person is left with a good value, the previous owners just have garbage data they can’t use anymore. If you created a thing on the heap and then gave it away, you wouldn’t have it anymore to free at the end. If you want to have “multiple owners” then you need ref counting and such, which also stops this problem of premature freeing.
Edit: one more thing: reference rules. You can have many read-only references to a thing, or one mutable reference. Unless you’re doing crazy things, the compiler simply won’t let you have references to a thing, and then via one of those references free that thing, thereby invalidating the other references.
Thats interresting, thanks! Stuff for me to look into!
I also think halfway through the conversation i might have given the impression i was talking about pointers, while it was not my intention to do so. That said, the readonly/mutable reference thing is very interresting!
Ill look into what rust does/has that is like the following psuedocode :
Rust simply doesn’t allow you to have references to data that goes out of scope (unless previously mentioned hoops are jumped through such as an explicitly declared unsafe block). It’s checked at compile time. You will never be able to compile the program.
Rust isn’t C. Rust isn’t C++. The memory-safe-ness of it is also not magic, it’s a series of checks in the compiler.
That sounds odd. That also means that a mapper, command, service,… can never return a class object or entity. Most of the programming world is based on oop o.O
Keep in mind im not talking about the usage of pointers, but reference typed variables.
The way I understand it, it is a bug in C implementation of free() that causes it to do something weird when you call it twice on the same memory. Maybe In Rust you can never call free twice, so you would never come across this bug. But, also Rust probably doesn’t have the same bug.
My point is it seems it is a bug in the underlying implementation of free(), not to be caught by the compiler, and can’t Rust have such errors no matter its superior design?
The way that rust attempts to prevent this class of error is not by making an implementation of free that is safe to call twice, but by making the compiler refuse to compile programs where free could be called twice on a pointer.
Anyway, use after free doesn’t depend on a double free. It just means that the program frees memory but keeps the pointer (which now points at memory that could contain unrelated data at some future point in time) and if someone trying to exploit the program finds a way to induce the program to read or write to that memory they may be able to access data they are not expected to, or write data to be used by a different part of the program that they shouldn’t be able to
Thanks, I understand the problem with using memory after it’s been freed and possibly access it changed by another part of the process. I guess I was confused by the double free explanation I read, which didn’t really say how it could be exploited, but I think you are right it still needs to be accessed later by the original program, which would not happen in Rust.
Not really, the issue is that C/C++ is not memory safe, i.e. it allows you to access memory that has already been freed. Consider the following C++ code:
int* wrong() { int data = 10; return &data; }If you try to use it it looks correct:
int* ptr = wrong(); std::cout << *ptr << std::endl;That will print
10, but the memory where data was defined has been freed, and is no longer in control of the program. Meaning that if something else allocated that memory they can control what my program does.Consider that on that example above later in the program we do:
If someone manages to get control of that memory between when we freed it and we used it they can make the access_level of the user be whatever they want.
This is a problem with C/C++ allowing you to access memory that has been freed, which is why C/C++ programmers need to be extra careful.
Thank you, that is very clear.